Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume that you have any two real numbers $a$ and $b$, and $1\leq k <\infty$, $k \in \mathbb{R}$.

How would you prove the inequality $|a+b|^k \leq 2^{k-1} (|a|^k+|b|^k)$?

share|improve this question
    
What about $k$? Is it a positive integer? A positive real? –  Patrick Da Silva Oct 17 '12 at 7:16
    
Edited, $1 \leq k < \infty$. –  Rojas Azules Oct 17 '12 at 7:17
1  
Sorry, $k \in \mathbb{R}$. –  Rojas Azules Oct 17 '12 at 7:20
add comment

1 Answer 1

up vote 3 down vote accepted

Method $1$:

The function $f(x) = x^k$ for $x\geq 0$ is a convex function for $k \geq 1$. Now apply Jensen's inequality.

Method $2$:

Let $\lvert a \rvert \geq \lvert b \rvert$. Let $t = \dfrac{b}a \implies 0 \leq \vert t \vert \leq 1$. We then want to prove that $$\vert 1 + t \vert^k \leq 2^{k-1} \left(1+\vert t \vert^k \right)$$ $$\left \vert \dfrac{1+t}2 \right \vert^k \leq \dfrac{1+\left \vert t \right \vert^k}2 $$ $$\left \vert \dfrac{1+t}2 \right \vert^k \leq \left(\dfrac{1+ \vert t \vert}{2} \right)^k$$ Setting $y = \vert t \vert \in [0,1]$, we want to prove that $$\left(\dfrac{1+ y}{2} \right)^k \leq \dfrac{1+y^k}2$$ $$f(y) = \dfrac{1+y^k}2 - \left(\dfrac{1+ y}{2} \right)^k$$ $$f'(y) = \dfrac{ky^{k-1}}2 - \dfrac{k}{2^k}(1+y)^{k-1} = \dfrac{k}2 \left(y^{k-1} - \left(\dfrac{1+y}2 \right)^{k-1} \right)$$ For $y \in [0,1]$, $y \leq \dfrac{1+y}2$, and hence $y^k \leq \left( \dfrac{1+y}2 \right)^k$. This means that $f'(y) < 0$ and hence $f(y)$ is decreasing. Hence, $$f(y) > f(1)$$ which implies $$\left(\dfrac{1+ y}{2} \right)^k \leq \dfrac{1+y^k}2$$

share|improve this answer
1  
I was just about to write exactly this. Haha –  Patrick Da Silva Oct 17 '12 at 7:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.